What is the slope of the line tangent to $f(x) = x^{2}-4x-3$ at $x = 3$ ?
Explanation: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{((x+h)^{2}-4(x+h)-3) - (x^{2}-4x-3)}{h}$ $ = \lim_{h \to 0} \frac{(x^{2}+2x h+h^{2}-4(x+h)-3) - (x^{2}-4x-3)}{h}$ $ = \lim_{h \to 0} \frac{x^{2}+2(x h)+h^{2}-4x-4h-3-x^{2}+4x+3}{h}$ $ = \lim_{h \to 0} \frac{2(x h)+h^{2}-4h}{h}$ $ = \lim_{h \to 0} 2x+h-4$ $ = 2x-4$ $ = (2)(3)-4$ $ = 2$